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%\title{Python 数学实验与建模} 
\title{4-黎曼-斯蒂尔杰斯积分}
%(1.1-1.2) 
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2023年1月6日} }

\maketitle

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%\begin{frame}[fragile=singleslide]{1.1.1. }
\begin{frame}{内容提要 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}
\item[4.1.]  黎曼积分
\item[4.2.]  黎曼-斯蒂尔杰斯积分 
\item[4.7.]  黎曼-斯蒂尔杰斯积分存在的一个充分条件
\item[4.8.]  黎曼-斯蒂尔杰斯积分的几个例子
\item[4.E.]  黎曼-斯蒂尔杰斯积分的一些练习题
\end{enumerate}

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\begin{frame}{4.1. 黎曼积分 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{enumerate}%\itemsep1em
\item  设有定义在区间 $[0,1]$ 上的实值函数 $f:[0,1]\to\mathbb{R}$. 
\item  设有分划 $\tau_n: 0 = t_0 < t_1 < t_2 < \cdots < t_n = 1$. 
\item  记 $\Delta_i=t_i-t_{i-1}$. 记 $\text{mesh}(\tau_n) = \underset{1\le i\le n}{\max}\Delta_i$. 
\item  设每个小区间取一点 $y_i \in [t_{i-1},t_i]$, 记 $\sigma_n=(y_1,y_2,\cdots,y_n)$. 
\item  定义黎曼和 $$ S_n = S_n(\tau_n, \sigma_n) = \sum\limits_{i=1}^{n} f(y_i)\Delta_i. $$
\item  如果当 $\text{mesh}(\tau_n)\to 0$ 时，$S_n(\tau_n,\sigma_n)$ 存在极限 $S$, 且该极限与 $\sigma_n$ 无关，那么称 $f$ 是一个黎曼可积函数，这个极限存为 $f$ 的黎曼积分，记为 $$S = \int_{0}^{1} f(t)dt.$$
\end{enumerate}

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\begin{frame}{4.2. 黎曼积分的例子 }

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\begin{itemize}%\itemsep1em

\item  问题：按定义计算黎曼积分 $\int_0^1 tdt. $
%\item  解答：
\begin{enumerate}
\item  将区间等分成 $n$ 份，即设 $t_i=i/n, 0\le i\le n$. 
\item  在每个小区间若取左端点则 $y_i=(i-1)/n$. 若取右端点则 $z_i=i/n$. 
\item  计算两个黎曼和
\begin{eqnarray*}
S(\tau_n,\sigma_n(y)) &=& \sum\limits_{i=1}^{n} \frac{i-1}{n}\cdot \frac{1}{n} = \frac{1}{n^2} \frac{n(n-1)}{2} \\
S(\tau_n,\sigma_n(z)) &=& \sum\limits_{i=1}^{n} \frac{i}{n}\cdot \frac{1}{n} = \frac{1}{n^2} \frac{n(n+1)}{2}. 
\end{eqnarray*}
\item  可见所有黎曼和的极限都是 $1/2$. 
\end{enumerate}

\end{itemize}

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\begin{frame}{4.3. 黎曼-斯蒂尔杰斯积分的概念 }

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\begin{enumerate}%\itemsep1em
\item  设有定义在区间 $[0,1]$ 上的实值函数 $f(t)$ 和实值单调递增函数 $g(t)$. 
\item  设有分划 $\tau_n: 0 = t_0 < t_1 < t_2 < \cdots < t_n = 1$. 
\item  记 $\Delta_i = t_i - t_{i-1}$. 记 $\text{mesh}(\tau_n) = \underset{1\le i\le n}{\max}\Delta_i$. 记 $\Delta_ig = g(t_i) - g(t_{i-1})$. 
\item  设每个小区间取一点 $y_i \in [t_{i-1},t_i]$, 记 $\sigma_n=(y_1,y_2,\cdots,y_n)$. 
\item  定义黎曼和 
\vspace{-0.5cm}
\begin{eqnarray*}
S_n = S_n(\tau_n, \sigma_n) = \sum\limits_{i=1}^{n} f(y_i)\Delta_ig. 
\end{eqnarray*}
\item  如果当 $\text{mesh}(\tau_n)\to 0$ 时，$S_n(\tau_n,\sigma_n)$ 存在极限 $S$, 且该极限与 $\sigma_n$ 无关，那么称 $f(t)$ 关于 $g(t)$ 是一个黎曼-斯蒂尔杰斯可积函数，记为 $$S = \int_{0}^{1} f(t)dg(t).$$
\end{enumerate}


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\begin{frame}{4.4. 黎曼-斯蒂尔杰斯积分的例子：分布函数  }

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\begin{itemize}\itemsep1em

\item  设 $F(x)$ 是随机变量 $X$ 的分布函数。则有如下两个黎曼-斯蒂尔杰斯积分
\begin{eqnarray*}
\int_{-\infty}^{\infty} dF(x) =1, \hspace{1cm} \int_{-\infty}^{\infty} xdF(x) = \mathbb{E}(X). 
\end{eqnarray*}

\item  具体例子：设离散型随机变量 $X$ 的分布律为 $\mathbb{P}\{X=0\}={1}/{3}$, $\mathbb{P}\{X=1\}={2}/{3}$.  则分布函数为
%\vspace{-0.1cm}
\begin{eqnarray*}
F(x) = \left\{\begin{array}{ll}
0, & x<0, \\
1/3, & 0\le x<1, \\
1, & 1\le x.
\end{array}\right.
\end{eqnarray*}
{\color{red}可以将 $dF(x)$ 理解为实数轴上的概率测度，包含 $x=0$ 的小区间的测度为 $1/3$, 包含 $x=1$ 的小区间的测度为 $2/3$. 不包含这两个点的任何区间的测度为零。}于是上述右边的积分等于 $0\cdot (1/3) + 1\cdot (2/3)=2/3$. 

\end{itemize}

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\begin{frame}{4.5. 上一页例子的概率函数和分布函数 }

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\begin{center}
\includegraphics[height=0.6\textheight, width=0.9\textwidth]{riemann-stieltjes.png}
\end{center}

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\begin{frame}{4.6. 函数的变差的概念 }

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\begin{itemize}%\itemsep1em

\item  {\color{red}The real-valued function $h$ on $[0,1]$ is said to have bounded $p$-variation for some $p>0$ if 
$$\sup\limits_{\tau} \sum\limits_{i=1}^{n} |h(t_i) - h(t_{i-1})|^p <\infty, $$
where the supremum is taken over all partitions $\tau$ of $[0,1]$. }

\item  例子：计算 $h(x)=\sin(x)$ 在 $[0,2\pi]$ 上的 $p$-变差，设 $p=1,2$. 
\item  例子：设 $h(x),x\in [0,1]$ 是布朗运动的一条样本路径。则当 $p>2$ 时 $h(x)$ 的 $p$-变差是有界的；当 $p\le 2$ 时，$h(x)$ 的 $p$-变差是无界的。

\end{itemize}

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\begin{frame}{4.7. 黎曼-斯蒂尔杰斯积分存在的一个充分条件 }

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\begin{itemize}\itemsep1em

\item  {\color{red}当定义于区间 $[0,1]$ 上的实值函数 $f(x)$ 和 $g(x)$ 满足下述两个条件的时候，黎曼-斯蒂尔杰斯积分  $$\int_0^1 f(x)dg(x)$$ 是存在的。}
\begin{enumerate}
\item  函数 $f(x)$ 和 $g(x)$ 在区间 $[0,1]$ 上没有共同的不连续点。
\item  存在正实数 $p$ 与 $q$ 使得 $p^{-1}+q^{-1}>1$, 而且 $f(x)$ 的 $p$-变差是有界的，$g(x)$ 的 $q$-变差是有界的。 
\end{enumerate}

\end{itemize}

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\begin{itemize}\itemsep1em

\item  设 $f(x)$ 是一个定义于区间 $[0,1]$ 上的可微函数。设 $B_t(\omega), t\in [0,1]$ 是布朗运动的一条样本路径。则下述黎曼-斯蒂尔杰斯积分是存在的，$$\int_0^1 f(t)dB_t(\omega)$$ 
\item  证明：可微函数的变差是有界的。即可取 $p=1$, $q>2$, 则上一页的充分条件成立。

\item  具体例子：下述黎曼-斯蒂尔杰斯积分都是存在的，
$$\int_0^1 e^tdB_t(\omega), \hspace{0.5cm} \int_0^1 \sin(t)dB_t(\omega), \hspace{0.5cm} \int_0^1 t^2dB_t(\omega). $$

\end{itemize}

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\begin{itemize}\itemsep1em

\item 设 $B_t(\omega), t\in [0,1]$ 是布朗运动的一条样本路径。则下述黎曼-斯蒂尔杰斯积分是不存在的，
$$\int_0^1 B_t(\omega)dB_t(\omega).$$

\item  验证1：当 $p>2$ 时，布朗运动的样本路径的 $p$-变差是有界的。在上述积分中， $p^{-1}+q^{-1}=2p^{-1}<1$. 因此前面的充分条件不成立。

\item  验证2：在分划 $\tau_n = (t_0,t_1,\cdots,t_n)$ 中的每个小区间 $[t_{i-1},t_i]$ 上，取左端点与取右端点形成的两个和式，收敛到不同的值。

\end{itemize}

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\begin{itemize}
\item  %第1题：单项选择
设随机变量 $X$ 的分布函数如下，求下述概率的值，和黎曼-斯蒂尔杰斯积分的值。
$$F(x)=\left\{ \begin{array}{ll} 0, & x<1, \\ 1/3, & 1\le x<3, \\ 1, & x\ge 3. \end{array} \right. \,\,\,  a=\mathbb{P}(X=1),\,\,\, b=\int_0^2 xdF(x). $$ 

\begin{enumerate}
\item[a.]  $a=1/3,\,\,\, b=1/3$. 
\item[b.]  $a=1/3,\,\,\,  b=2/3$. 
\item[c.]  $a=2/3,\,\,\,  b=1/3$. 
\item[d.]  $a=2/3,\,\,\,  b=2/3$. 

\end{enumerate} 

\vspace{0.2cm}

\end{itemize}

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\begin{itemize}

\item  {\color{red}解答：b. 根据分布函数可知这个随机变量是离散型的，并且事件 $X=1$ 的概率是 $1/3$, 事件 $X=3$ 的概率是 $2/3$. 这个黎曼-斯蒂尔杰斯积分，概率测度 $dF(x)$ 在包含 $x=1$ 的小区间上的测度是 $1/3$, 在包含 $x=3$ 的小区间上的测度是 $2/3$. 因此这个积分的值是 $2/3$.  

}

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\begin{itemize}
\item  %第2题：单项选择
实值函数 $h(x), 0\le x\le 1$ 的 $p$-变差定义为上确界 
$$\sup\limits_{\tau} \sum\limits_{i=1}^{n} |h(t_i) - h(t_{i-1})|^p, $$
其中的 $\tau=(t_0,t_1,\cdots,t_n)$ 取遍这个区间的所有有限分划。计算函数 $h(x)=x, 0\le x\le 1$ 的 1-变差 $a$ 与 2-变差 $b$. 
 
\begin{enumerate}
\item[a.]  $a=1,\,\,\, b=1$. 
\item[b.]  $a=0,\,\,\, b=1$. 
\item[c.]  $a=1,\,\,\, b=0$. 
\item[d.]  $a=0,\,\,\, b=0$. 

\end{enumerate} 

\vspace{0.2cm}

\end{itemize}

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\item  {\color{red}解答：c. 因为这个函数是单调递增的，所以 1-变差为 $h(1)-h(0)=1$. 现在计算 2-变差。考虑分划 
$$\tau:\,\,\, 0<\frac{1}{n}<\frac{2}{n}<\cdots<\frac{n-1}{n}<1. $$ 则可以知道从这个分划得到的 2变差为 $$\sum\limits_{k=1}^n (1/n)^2 = 1/n,$$ 当 $n\to\infty$ 时，这个极限是 0. 因此猜测 2-变差是零。但是严格怎么证呢？

}

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\begin{thebibliography}{99}

\bibitem{mikosch} {\ppr Thomas Mikosch}. {\ppr Elementary Stochastic Calculus}. 世界图书出版公司，{\ppr 2009} 年 {\ppr 8} 月第 {\ppr 1} 版。
\bibitem{wangjun} 王军、邵吉光、王娟. 随机过程及其在金融领域中的应用. 清华大学出版社，北京交通大学出版社，{\ppr 2018} 年{\ppr 8} 月第 {\ppr 2} 版。
\bibitem{zhangbo} 张波、商豪. 应用随机过程. 中国人民大学出版社，{\ppr 2016} 年 {\ppr 6} 月第 {\ppr 4} 版。
%\bibitem{karlin} {\ppr Mark A. Pinsky, Samuel Karlin}. {\ppr An Introduction to Stochastic Modeling}. 机械工业出版社，{\ppr 2013} 年 {\ppr 2} 月第 {\ppr 1} 版。


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